Solution:
11) Let us define some events as follows:
A : A resident has a privacy fence
B : A resident has a dog
a) We have to obtain P(A|B).
Conditional probability of A given B is given as follows:
P(A ∩ B) = Probability that a resident own a dog and also he has privacy fence
P(B) = Probability that a resident own a dog
We have, P(A ∩ B) = 25% = 0.25 and P(B) = 35% = 0.35
Hence, P(A|B) = 0.25/0.35 = 5/7
Given that a resident own a dog, the probability that he has a privacy fence is 5/7.
b) If two events A and B are independent then, P(A ∩ B) = P(A).P(B)
We have, P(A) = 0.60, P(B) = 0.35 and P(A ∩ B) = 0.25
Now, P(A).P(B) = 0.60×0.35 = 0.21 which is not equal to P(A ∩ B).
Hence, having a dog and having a fence are not independent events.
c) If two events A and B are disjoint events then P(A ∩ B) = 0
We are given that, P(A ∩ B) = 0.25 which is not equal to zero.
Therefore, having a dog and having a privacy fence are not disjoint events.
11) The probability of an Emerald Gates subdivision resident having a privacy fence is 60%. The...
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